Question

A solid sphere is rolling. What is the fraction of rotational kinetic energy associated with it to its total kinetic energy ?

• A. $2/7$
• B. $5/7$
• C. $1/2$
• D. $3/4$

Answer 1

Answer: (A)

Rotational kinetic energy $K_R=\frac{1}{2}I{\omega }^2$
For a solid sphere $I=\frac{2}{5}M{R}^2$
$\therefore K_R=\frac{1}{2}\times \frac{2}{5}M{R}^2{\omega }^2$
Translational kinetic energy $K_T=\frac{1}{2}M{v}^2$
Total kinetic energy $K=K_R+K_T$
$K=\frac{1}{5}M{R}^2{\omega }^2+\frac{1}{2}M{v}^2$
$=\frac{1}{5}M{R}^2{\omega }^2+\frac{1}{2}M{R}^2{\omega }^2$
$(\because v=R\omega )$
$=\frac{7}{10}M{R}^2{\omega }^2$
$\therefore \frac{\text{Kinetic energy of rotation}{K}_R}{\text{Total energy K }}$
$=\frac{\frac{1}{5}M{R}^2{\omega }^2}{\frac{7}{10}M{R}^2{\omega }^2}$
$=\frac{2}{7}$